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Question Number 141885 by mnjuly1970 last updated on 24/May/21

easy question: if lim_(x→0) ((1−cos(1−cos(1−cos(x))))/x^8 ) =2^( a) then a=??

$$\:\:\: \\ $$$$\:\:\:\:\:\:{easy}\:\:{question}: \\ $$$$\:\:\:\:{if}\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)}{{x}^{\mathrm{8}} }\:=\mathrm{2}^{\:{a}} \\ $$$$\:\:\:\:\:\:\:\:{then}\:\:\:{a}=??\:\: \\ $$

Answered by Dwaipayan Shikari last updated on 24/May/21

lim_(x→0) ((1−cos(1−cos(1−cos(x))))/x^8 )=((1−cos(1−cos(x^2 /2)))/x^8 )=((1−cos(2sin^2 (x^2 /4)))/x^8 ) =((1−cos((x^4 /8)))/x^8 )=((2sin^2 ((x^4 /(16))))/x^8 )=(1/(128))=2^(−7)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)}{{x}^{\mathrm{8}} }=\frac{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{{x}^{\mathrm{8}} }=\frac{\mathrm{1}−{cos}\left(\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}^{\mathrm{2}} }{\mathrm{4}}\right)}{{x}^{\mathrm{8}} } \\ $$$$=\frac{\mathrm{1}−{cos}\left(\frac{{x}^{\mathrm{4}} }{\mathrm{8}}\right)}{{x}^{\mathrm{8}} }=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}^{\mathrm{4}} }{\mathrm{16}}\right)}{{x}^{\mathrm{8}} }=\frac{\mathrm{1}}{\mathrm{128}}=\mathrm{2}^{−\mathrm{7}} \\ $$

Commented by mnjuly1970 last updated on 24/May/21

thanks alot..

$${thanks}\:{alot}.. \\ $$

Answered by iloveisrael last updated on 24/May/21

lim_(x→0) ((2sin^2 (((1−cos (1−cos x))/2)))/x^8 ) = lim_(x→0) ((2sin^2 (((2sin^2 (((1−cos x)/2)))/2)))/x^8 ) = lim_(x→0) ((2sin^2 (((2sin^2 (((2sin^2 ((x/2)))/2)))/2)))/x^8 ) = lim_(x→0) ((2.sin^2 ((x^4 /2^4 )))/x^8 ) = lim_(x→0) ((2.((x^8 /2^8 )))/x^8 ) = (1/2^7 )

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{\mathrm{2}}\right)}{{x}^{\mathrm{8}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{2}}\right)}{\mathrm{2}}\right)}{{x}^{\mathrm{8}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\cancel{\mathrm{2}sin}\:^{\mathrm{2}} \left(\frac{\cancel{\mathrm{2}sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\cancel{\mathrm{2}}}\right)}{\cancel{\mathrm{2}}}\right)}{{x}^{\mathrm{8}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}.\mathrm{sin}\:^{\mathrm{2}} \left(\frac{{x}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} }\right)}{{x}^{\mathrm{8}} }\: \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}.\left(\frac{{x}^{\mathrm{8}} }{\mathrm{2}^{\mathrm{8}} }\right)}{{x}^{\mathrm{8}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{7}} } \\ $$

Answered by mathmax by abdo last updated on 24/May/21

1−cosx∼(x^2 /2) ⇒cos(1−cosx)∼cos((x^2 /2))∼1−(x^4 /8) ⇒ 1−cos(1−cosx)∼(x^4 /8) ⇒cos(1−cos(1−cosx))∼cos((x^4 /8)) ∼1−(x^8 /(2.64)) ⇒1−cos(1−cos(1−cosx))∼(x^8 /2^7 ) ⇒ lim_(x→0) ((1−cos(1−cos(1−cosx)))/x^8 )=(1/(128))=2^(−7) ⇒a=−7

$$\mathrm{1}−\mathrm{cosx}\sim\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{cos}\left(\mathrm{1}−\mathrm{cosx}\right)\sim\mathrm{cos}\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{8}}\:\Rightarrow \\ $$$$\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cosx}\right)\sim\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{8}}\:\Rightarrow\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cosx}\right)\right)\sim\mathrm{cos}\left(\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{8}}\right) \\ $$$$\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{2}.\mathrm{64}}\:\Rightarrow\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cosx}\right)\right)\sim\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{2}^{\mathrm{7}} }\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cosx}\right)\right)}{\mathrm{x}^{\mathrm{8}} }=\frac{\mathrm{1}}{\mathrm{128}}=\mathrm{2}^{−\mathrm{7}} \:\Rightarrow\mathrm{a}=−\mathrm{7} \\ $$

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